Question: The base of a solid is the region enclosed by the graphs of $y=\text{sin}(x)$ and $y=4-\sqrt x$, between $x=2$ and $x=7$. Cross sections of the solid perpendicular to the $x$ -axis are rectangles whose height is $2x$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_2^7 [\text{sin}(x)+\sqrt x-4]\cdot 2x\,dx$ (Choice B) B $\int_2^7 [(4-\sqrt x)^2-\text{sin}^2(x)]\,dx$ (Choice C) C $\int_2^7 [4-\sqrt x-\text{sin}(x)]\cdot 2x\,dx$ (Choice D) D $\int_2^7 [\text{sin}^2(x)-(4-\sqrt x)^2]\,dx$
Explanation: It's always a good idea to graph the situation before we solve. $y$ $x$ $ R$ ${y=\text{sin}(x)}$ ${y=4-\sqrt{x}}$ $ 2$ $ 7$ Let's imagine the solid is made out of many thin slices. $y$ $x$ Each slice is a prism. Let the width of each slice be $dx$ and let the area of the prism's face, as a function of $x$, be $A(x)$. Then, the volume of each slice is $A(x)\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small width using a definite integral: $\int_a^b A(x)\,dx$ What we now need is to figure out the expression of $A(x)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=\text{sin}(x)}$ ${y=4-\sqrt{x}}$ $ 2$ $ 7$ $ b(x)$ $ h(x)$ $ dx$ $ A(x)$ The face of that slice is a rectangle with base $b(x)$ and height $h(x)$. We are given that $h(x)=2x$, and we know that for each value of $x$, the base $b(x)$ is equal to the difference between ${y=4-\sqrt x}$ and ${y=\text{sin}(x)}$. Now we can find an expression for the area of the face of the prism: $\begin{aligned} &\phantom{=}A(x) \\\\ &=b(x)h(x) \\\\ &=[({4-\sqrt{x}})-{\text{sin}(x)}]\cdot 2x \\\\ &=[4-\sqrt x-\text{sin}(x)]\cdot 2x \end{aligned}$ The leftmost endpoint of the base of the solid is at $x=2$ and the rightmost endpoint is at $x=7$. So the interval of integration is $[2,7]$. Now we can express the definite integral in its entirety! $\int_2^7 [4-\sqrt x-\text{sin}(x)]\cdot 2x\,dx$